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$V$ is an $n$-dimensional vector space. Show that $n + 1$ vectors in $V$ form a linearly dependent set.

Here is how I am approaching it:

Let $\dim V = n$, which implies that $S$ is a linearly independent set of vectors such that $S = \{v_1,v_2,\ldots,v_n\}$ is the basis of $V$.

Let $W = \{w_1,\ldots,w_r\}$ be a set of linearly independent vectors in $V$

I don't know where to go from here.

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    $\begingroup$ You need to use a basis for $V$. $V$ is not a set with $n$ vectors. It has infinitely many vectors. But if its dimension is $n$ then you can pick up a basis with $n$ LI vectors. $\endgroup$ – Sigur Nov 2 '13 at 16:32
  • $\begingroup$ @Sigur Updated it to what I think your comment was leading me towards. Do you have any advice on where to go from here? $\endgroup$ – n8sty Nov 2 '13 at 16:40
  • $\begingroup$ Can you assume that n of them are base, then imply that the (n+1)th vector can be displayed as linear combination of the others? $\endgroup$ – LeeNeverGup Nov 2 '13 at 16:45
  • $\begingroup$ @LeeNeverGup That's what the question requires as an answer. I don't know how to show that. $\endgroup$ – n8sty Nov 2 '13 at 16:48
  • $\begingroup$ It's better to use TeX syntax to type math. $\endgroup$ – leo Nov 2 '13 at 18:04
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The statement $\dim V = n$ means there exists a basis $\{v_1,\ldots,v_n\}$ for $V$. Let $v_{n+1} \in V\setminus \{v_1,\ldots,v_n\}$. Then by the definition of a basis, there exist $\alpha_1,\ldots,\alpha_n \in F$ ($F$ being the field over which $V$ is defined) such that $$ v_{n+1} = \alpha_1v_1 + \ldots + \alpha_nv_n. $$ So $\{v_1,\ldots,v_{n+1}\}$ is linearly dependent.

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    $\begingroup$ This doesn't prove "any" $n+1$ vectors is LD. $\endgroup$ – user85798 Oct 25 '14 at 23:44
  • $\begingroup$ @LTS: As I recall, my intent with the answer was to let the OP reach the desired conclusion based on the above, since it follows almost immediately. Indeed, since we are assuming $V$ is of dimension $n$, every set of $n$ linearly independent vectors in $V$ is a basis for $V$. If $S \subset V$ is of order $n+1$, then either every $n$ element subset of $S$ is linearly dependent (in which case so is $S$) or $S$ contains a basis for $V$. The set of vectors is linearly dependent in the former case immediately and in the latter case based on my answer. $\endgroup$ – Dan Oct 26 '14 at 19:53
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Believe I can answer my own question:

Assuming the above let Let $W_1$ = {$w_1$, $v_1$, $v_2$, . . . $v_n$}

$W_1$ spans V because S spans V.

$w_1$ is a linear combination of the vectors in S because S is a basis for V.

Then we get $w_1$ = $a_1$$v_1$ + $a_2$$v_2$ + . . . + $a_n$$v_n$ rearrange terms so that 0 = $a_1$$v_1$ + $a_2$$v_2$ + . . . + $a_n$$v_n$ - $w_1$ and $W_1$ is a linearly dependent set because at least one of the coefficients does not equal 0/

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